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STRUCTURE OF CERIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( September 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Naturally occurring cerium (Ce) is composed of 4 stable isotopes: 136Ce, 138Ce, 140Ce, and 142Ce with 140Ce being the most abundant (88.48% natural abundance) and the only one theoretically stable; 136Ce, 138Ce, and 142Ce are predicted to undergo double beta decay but this process has never been observed. 35 radioisotopes have been characterized with the most stable being 144Ce with a half-life of 284.893 days, 139Ce with a half-life of 137.640 days, and 141Ce with a half-life of 32.501 days. All of the remaining radioactive isotopes have half-lives that are less than 4 days and the majority of these have half-lives that are less than 10 minutes. Comparing the cerium of 58 protons (even number) with Barium of 56 protons (even number ) we conclude that the structure of cerium -116 has the same high symmetry of the structure of Barium-112. ( See my STRUCTURE OF Ba-132..Ba-138 ). Using the following diagram of Ce-116 based on the structure of Ba-112 we conclude that in Ce-116 with S = 0 the additional p57n57 and p58n58 with S=0 which are not shown contribute to the high symmetry with total S = 0. ' ' ' '''STRUCTURE OF Ce-120, Ce-122, Ce-124, Ce-126, Ce-128, Ce-130, Ce-132, Ce-134, Ce-134, Ce-136, Ce-138, Ce-140, AND Ce-142 WITH S = 0 ' For understanding the structure of the above nuclides having even number of extra neutrons you must read my STRUCTUTRE OF Ce-136 . Here we have an even number of extra neutrons based on the structure of Ce-116 with S = 0, because they appear with opposite spins and fill the blank positions of the structure of Ce-116 with S = 0. For example the unstable Ce-134 with S = 0 has 18 extra neutrons with opposite spins. These extra neutrons make two bonds per neutron but the small number of them cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable structures of Ce-136, Ce-138, Ce-140 and Ce-142 the greater number of extra neutrons gives enough binding energies to pn bonds for overcoming the repulsions. '''STRUCTURE OF Ce-144, Ce-146, Ce-148, Ce-150, Ce-152, Ce-154, AND Ce-156 WITH S = 0 Similarly the structure of the above unstable nuclides with even number of extra neutrons is based on the same structure of Ce-116 with S =0. For example the Ce-144 with S =0 has 28 extra neutrons of opposite spins but the two more extra neutrons than those of the stable Ce-142 (in the absence of blank positions ) make single bonds leading to the decay. In other words in the nuclides from C-146 to Ce-156 the more extra neutrons than those of the Ce-142 make single bonds leading to the decay. ' ' SRUCTURE OF Ce-133, Ce-135, Ce-137, Ce-139, Ce-143, Ce-145, Ce-149, Ce-151 AND Ce-153 After a careful analysis I found that the structure of the above unstable nuclides with odd number of extra neutrons is based also on the structure of Ce-116 with S =0 . For example the Ce-133 with S = +1/2 of 17 extra neutrons has 9 extra neutrons of positive spins and 8 extra neutrons of negative spins. That is S = 0 + 9(+1/2) + 8(-1/2) = +1/2 Whereas the Ce-153 with S =-3/2 of 37 extra neutrons has 17 extra neutrons of positive spins and 20 extra neutrons of negative spins. That is S = 0 + 17(+1/2) + 20(-1/2) = -3/2 STRUCTURE OF Ce-119, Ce-121, Ce-123, Ce-127, Ce-129, Ce-131, AND Ce-157 After a careful analysis I found that the structure of the above nuclides with such an odd number of extra neutrons is based on another structure of Ce-116 having S = +2 . Using the diagram of Ce-116 with S =0 we conclude that the p37n37 changes the spin from S=-1 to S =+1 giving S = +2. In this case it moves from the -HSQ to +HSQ for making horizontal bonds with p38n38. Then in the presence of odd number of extra neutrons we get the structures of the above nuclides. For example the Ce-119 with S +5/2 of 3 extra neutrons has two extra neutrons of positive spins and one extra neutron of negative spin. That is S = +2 + 2(+1/2) + 1(-1/2) = +5/2 . Whereas, the Ce-157 with +7/2 of 41 extra neutrons has 22 extra neutrons of positive spins and 19 extra neutrons of negative spins. That is S = +2 + 22(+1/2) + 19(-1/2) = +7/2 STRUCTURE OF Ce-125, Ce-141, Ce-147, AND Ce-155 Here the structure of the above nuclides with odd number of extra neutrons having negative spins is based on another structure of Ce-116 having S = -2 . In this case using the diagram of Ce-116 with S =0 we conclude that the p38n38 changes the spin from S=+1 to S =-1 giving S = -2. Particularly it moves from the +HSQ to -HSQ for making horizontal bonds with p39n39. Then in the presence of odd number of extra neutrons we get the structures of the above nuclides. For example the Ce-125 with S -7/2 of 9 extra neutrons has 4 extra neutrons of positive spins and 5 extra neutron of negative spins. That is S = -2 + 4(+1/2) + 5(-1/2) = -7/2 . ' ' ' DIAGRAM OF Ce-116 WITH S = 0 ' This structure of Ce-116 with S = 0 is based on Ba-112 with S= 0, because the additional p55n55 and p56n56 as vertical systems with S = 0 are not shown. Here you see the 6 horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4, +HP5 and -HP6 along the two horizontal squares, the -HSQ , and +HSQ . ' n40.......p40' ' +HSQ p38..........n38 ' ' n31………p12.........n12.......p32' ' -HP6 p31........n11.........p11…… n32 ' ' p29.........n10.........p10…… n30' ' +HP5 n29………p9........n9 …….p30' ' p47.......n27.........p8............n8.........p28........n48' ' -HP4 n45.......p27.........n7..........p7............n28........p46 ' ' n47......p25.........n6.........p6............n26.........p48' ' +HP3 p45......n25……….p5..........n5……...p26........n46 ' ' n23……p4..........n4………….p24' ' -HP2 p23……....n3………p3………..n24 ' ' p21.........n2………p2...........n22' ' +HP1 n21........p1........n1.........p22' ' p37......n37 ' ' -HSQ n39.....p39' ' ' Category:Fundamental physics concepts